Binomial probability distribution

See attached file

# 48. It is known that diskettes produced by a certain company will be defective with probability 0.01, independently of each other. The company sells the diskettes in packages of size 10 and offers a money-back guarantee that at most 1 of 10 diskettes in the package will be defective. If someone buys 3 packages, what is the probability that he or she will return exactly one of them?

This is binomial probability distribution
Probability of a defective diskette = p= 0.01
Therefore probability that the diskette is not defective= q= 1-p= 1-0.01=0.99
Probability that in a package of 10 diskettes will have r defective diskettes
= n C 4 p r q n-r
Here n= 10
Therefore Probability of r defectives = 10 C 4 p r q 10-r
Probability of 0 defectives (r=0)
= 10 C 0 p 0 q 10 = 1 x (0.01) 0 x (0.99) 10 = 0.9044
Probability of 1 defectives (r=1)
= 10 C 1 p 1 q 9 = 10 x (0.01) 1 x (0.99) 9 = 0.0914
A package is not considered defective if it has 0 or 1 defective diskette
Therefore probability of a non defective package = 0.9044 + 0.0914 = 0.9958
Probability of defective package + Probability of non defective package = 1
Or Probability of defective package =1- Probability of non defective package = 1-0.9958 = 0.0042

Now we have to calculate the probability that exactly one in 3 packages is defective and hence to be returned.
Again this is a binomial probability distribution
Probability of a defective package = p= 0.0042
Therefore probability that a package is not defective= q= 1-p= 1-0.0042=0.9958
Probability that in 3 packages exactly 1 package is defective
= 3 C 1 p 1 q 2
= 3 x (0.0042) 1 x (0.9958) 2 = 0.0125
Answer Probability that exactly 1 package will be returned= 0.0125 or 1.25%

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!